Hyperbolas Covered by a Triangle

Solution: Obviously, to be the smallest triangle, the vertices of the triangle must be on the hyperbolas, or in the interior area with edges tangents to the hyperbolas, or any combinations of the above two cases.

Otherwise, if any edge intersects a hyperbolas, or a vertex is outside of the interior area, a smaller triangle can be always found by moving the edge tangent to the hyperbola, or moving the vertex toward the interior region.

Case 1: There is only two vertices on two adjacent branches, and the third one in the interior with its two edges that tangent to the other two branches.

Step 1 – tangent line to $xy=k$
For $xy=k$, the tangent at $(x_0,y_0)$, with $x_0y_0=k$, is $$y_0x+x_0y=2k$$

Step 2 – find tangent parameters
For $y=−\dfrac{1}{x}$ (i.e. $xy=−1$), let the tangent point be $(t,−\dfrac{1}{t})$, where $t>0$. The tangent line is: $$−\dfrac{1}{t}x+ty=−2\tag{1}$$

This line must pass through $B=(−b,−\dfrac{1}{b})$, where $b>0$. Substituting gives $$\dfrac{b}{t}−\dfrac{t}{b}=−2$$

so $t$ satisfies $$t^2−2bt−b^2=0$$ The root with $t<0$ is $t=(1−\sqrt{2})b$. Let $\gamma=\sqrt{2}-1$, we have $t=-b\gamma$.

For $y=\dfrac{1}{x}$ (i.e. $xy=1$), let the tangent point be $(s,\dfrac{1}{s})$, where $s>0$. The tangent line is: $$\dfrac{1}{s}x+sy=2\tag{2}$$

This line must pass through $C=(c,−\dfrac{1}{c})$, where $c>0$. Substituting gives $$\dfrac{c}{s}-\dfrac{s}{c}=2$$

so $s$ satisfies $$s^2+2cs−c^2=0$$ The root with $s>0$ is $s=c(\sqrt{2}−1)=c\gamma$.

Step 3 – intersection $A=(u,v)$ of the two tangents
Solving the two linear tangent equations $(1)$ and $(2)$, for value of $(x,y)=(u,v)$ gives $$u=\dfrac{2st(s+t)}{s^2+t^2}\ \ \ \ \ \ \ v=\dfrac{2(s−t)}{s^2+t^2}$$

Simplifying the above with $s=c\gamma$, $t=-b\gamma$, we have $$u=\dfrac{2bc(b−c)\gamma}{b^2+c^2}\ \ \ \ \ \ v=\dfrac{2(b+c)}{(b^2+c^2)\gamma}$$

Step 4 – area of $\triangle{ABC}$
Use the determinant formula $$[\triangle{ABC}]=\dfrac{1}{2}∣det⁡([B−A,  C−A])∣$$

Carrying out the algebra, by substituting the A,B,C coordinates above, yields the closed form $$[\triangle{ABC}]=\dfrac{((b−c)^2\gamma+2bc)((b+c)^2+2bc\gamma)}{2bc(b^2+c^2)\gamma}$$

where $\gamma=\sqrt{2}−1$.

Replacing $\gamma=\sqrt{2}−1$ gives the following: $$[\triangle{ABC}]=\dfrac{((\sqrt{2}−1)(b−c)^2+2bc)((b+c)^2+2(\sqrt{2}−1)bc)}{2(\sqrt{2}-1) bc (b^2+c^2)}$$

Simplifying the above formula, we have $$[\triangle{ABC}]=\dfrac{(b^2+c^2+2\sqrt{2}bc)^2}{2bc(b^2+c^2)}$$

Step 5 – smallest area of $\triangle{ABC}$

Let $x=\dfrac{b}{c}$, where $x>0$, we have $$[\triangle{ABC}]=f(x)=\dfrac{(x^2+2\sqrt{2}x+1)^2}{2x(x^2+1)}$$

Because $$f'(x)=\dfrac{x^6-7x^4+7x^2-1}{2x^2(x^2+1)^2}=\dfrac{(x^2-1)(x^4-6x^2+1)}{2x^2(x^2+1)^2}$$

Let $f'(x)=0$, we find the critical points as $$x^2=1, \ \ \ \ \ \ x^2=3\pm2\sqrt{2}$$ Because $x>0$, we have $$x=1,\ \ \ \ \ \ x=\sqrt{2}-1,\ \ \ \ \ \ $x=\sqrt{2}+1$$

Evaluating $f(1)$, $f(\sqrt{2}-1)$ and $f(\sqrt{2}+1)$, we have $$f(1)=3+2\sqrt{2}\ \ \ \ \ \ f(\sqrt{2}-1)=4\sqrt{2}\ \ \ \ \ \ f(\sqrt{2}+1)=4\sqrt{2}$$

As $t\rightarrow\infty$, $f(x)\rightarrow \infty$, the minimum value of $[\triangle{ABC}]=4\sqrt{2}$ when $x=\dfrac{b}{c}=\sqrt{2}\pm 1$.

Case 2: All three vertices on three branches, and one edge tangents to the remaining branch, two vertices are the tangent points of other two edges

Let $A=(-a,\dfrac{1}{a})$, $B=(b, \dfrac{1}{b})$, $C=(-c,-\dfrac{1}{c})$, and $a,b,c>0$. The equation of line $BC$ is $$\dfrac{y-\frac{1}{b}}{-\dfrac{1}{c}-\dfrac{1}{b}}=\dfrac{x-b}{-c-b}$$ Simplifying it, we have $$y=\dfrac{1}{bc}x+\dfrac{1}{b}-\dfrac{1}{c}$$

Because $BC$ intersects to the branch in the 4th quadrant, we have $b>c$ and $$-\dfrac{1}{x}=\dfrac{1}{bc}x+\dfrac{1}{b}-\dfrac{1}{c}$$ i.e. $$x^2+(c-b)x+bc=0\tag{3}$$ Because $BC$ is tangent to the branch in the 4th quadrant, Equation (1) has only one root. Therefore $$\Delta=(c-b)^2-4bc=0$$ i.e $$b^2-6bc+c^2=0$$ Solving $b$ in the above equation, we have $$b=(3\pm2\sqrt{2})c$$ Given $b>c$, we have $$b=(3+2\sqrt{2})c$$

Use shoelace formula, $$[\triangle{ABC}]=\dfrac{1}{2}(x_ay_c-x_cy_a+x_cy_b-x_by_c+x_by_a-x_ay_b)$$ $$=\dfrac{1}{2}((-a)\cdot(-\dfrac{1}{c})-(-c)\cdot\dfrac{1}{a}+(-c)\cdot\dfrac{1}{b}-b\cdot(-\dfrac{1}{c})+b\cdot\dfrac{1}{a}-(-a)\cdot\dfrac{1}{b})$$ $$=\dfrac{1}{2}(\dfrac{a}{b}+\dfrac{b}{a}+\dfrac{c}{a}+\dfrac{a}{c}+\dfrac{b}{c}-\dfrac{c}{b})$$

Because $b=(3+2\sqrt{2})c$, we can simplify the above equation as $$[\triangle{ABC}]=(2+\sqrt{2})\dfrac{c}{a}+(2-\sqrt{2})\dfrac{a}{c}+2\sqrt{2}\tag{4}$$

By AM-GM inequality, we have $$(2+\sqrt{2})\dfrac{c}{a}+(2-\sqrt{2})\dfrac{a}{c}\ge 2\sqrt{(2+\sqrt{2})\dfrac{c}{a}\cdot(2-\sqrt{2})\dfrac{a}{c}}\ge 2\sqrt{2}\tag{5}$$

The equality holds when $$(2+\sqrt{2})\dfrac{c}{a}=(2-\sqrt{2})\dfrac{a}{c}$$

Because $a,b,c>0$, solve the above equation, we have $$a=(\sqrt{2}+1)c$$

Therefore, combining $(4)$ and $(5)$, the minimum value of $[\triangle{ABC}]$ is $4\sqrt{2}$, when $$a=(\sqrt{2}+1)c$$ $$b=(3+2\sqrt{2})c$$ or $$b=(\sqrt{2}+1)a$$ $$c=(\sqrt{2}-1)a$$

which is a covered by Case 1.

Therefore, the smallest area of triangles intersecting all 4 hyperbola branches of $x^2y^2=1$ is $\boxed{4\sqrt{2}}$.